Core Pure 2

Question 1

1.1 What is the exponential form of a complex number?

Answer 1

z = re^(iθ)

where r is mod(z) and θ is arg(z)

Question 2

1.1 What is Euler’s relation?

Answer 2

e^(iθ) = cosθ + isinθ

Question 3

1.1 How do you convert from a complex number in exponential form to cartesian form?

Answer 3

r is the modulus so r = sqrt(x^2 + y^2)

θ is the argument so tanθ = y/x

Solve simultaneously, using a diagram to determine +ve or -ve signs

Question 4

1.2 How do you multiply two complex numbers, z1=r1e^(iθ1) and z2=r2e^(iθ2) in exponential form?

Answer 4

z1z2 = r1r2e^i(θ1 + θ2)

Question 5

1.2 How do you divide two complex numbers, z1=r1e^(iθ1) and z2=r2e^(iθ2) in exponential form?

Answer 5

z1/z2 = (r1/r2)e^i(θ1 - θ2)

Question 6

1.3 What is De Moivre’s theorem?

Answer 6

For any integer n,

(r(cosθ + isinθ))^n = r^n(cos(nθ) + isin(nθ))

Question 7

1.4 What are some trigonometric rules you can apply when solving De Moivre’s theorem questions?

Answer 7

z + 1/z = 2cosθ

z - 1/z = 2isinθ

z^n + 1/z^n = 2cosnθ

z^n - 1/z^n = 2isinnθ

Question 8

1.4 How do you express a trig function as powers of trig functions using De Moivre’s theorem?

Answer 8

cosnθ or sinnθ:
(cosθ + isinθ)^n = cosnθ + isinnθ

Expand LHS binomially, equate real parts if cos or imaginary parts if sin (and then divide by i)

Question 9

1.4 How do you express a trig function to a power as trig functions using De Moivre’s theorem?

Answer 9

2cosθ = z + 1/z
(2cosθ)^n = (z + 1/z)^n

Expand both sides, using binomial expansion on RHS, then rearrange to find cos^nθ on its own

If finding sin then use the same method but use z - 1/z instead and 2isinθ on the LHS, be wary of any i’s

Question 10

1.5 What is the formula for the sum of a geometric series with complex numbers?

Answer 10

Σ(0->n-1) wz^r = w + wz + wz^2 + … + wz^(n-1) = (w(z^(n)-1)/(z-1)

where w is the starting number and z is the common ratio

Question 11

1.5 What is the formula for the sum of an infinite geometric series with complex numbers?

Answer 11

Σ(0->∞) wz^r = w + wz + wz^2 +… = w/(1-z), where mod(z) < 1 (convergent series)

where w is the starting number and z is the common ratio

Question 12

1.5 How is the formula of the sum of a geometric sequence related to the values of the sigma notation?

Answer 12

If the value at the top of the sigma notation is k, then the numerator of the formula will be (z^(k+1) -1)

Question 13

1.5 When working with sums of geometric series with complex numbers, what rules must you look out for?

Answer 13

i^2 = -1 but also -1 can be written as i^2 (useful for getting i off of the denominator)

e^iθ -1 = e^(iθ/2) (e^(iθ/2) - e^(-iθ/2) )
and (e^(iθ/2) - e^(-iθ/2) ) can be written as 2isinθ (or 2isin(nθ) if the power has an n in it)

Similarly, (e^(iθ/2) + e^(-iθ/2) ) can be written as 2cosθ (or 2cosnθ if the power has an n in it)

e^πi = -1 or e^πi + 1 = 0 (Euler’s Identity)

Question 14

1.6 If z and w are non-zero complex numbers, and n is a positive integer, how many distinct solutions does z^n = w have?

Answer 14

n distinct solutions

Question 15

1.6 How do you find the nth roots of a complex number?

Answer 15

z^n = r(cosθ + isinθ)
z^n = r(cos(θ + 2kπ) + isin(θ + 2kπ))
z=r^(1/n) (cos((θ + 2kπ)/6)) + isin((θ + 2kπ)/6))

Substitute in n values of k, starting from k=0, to find solutions in the range -π<θ≤π, the solutions should all the same argument of difference between them, e.g. if n=4, each argument is π/2 more than the last

Question 16

1.6 How do you write 1 in modulus argument form?

Answer 16

cos0 + isin0

Next step is:
cos(0+2kπ) + isin(0+2kπ)
= cos(2kπ) + isin(2kπ)

Question 17

1.7 What do the nth roots of a complex number form on n Argand Diagram?

Answer 17

nth roots of a complex number lie at the vertices of a regular n-gon with its centre at the origin

Question 18

1.7 How can you find the vertices of the regular polygon formed by the nth roots of a complex number?

Answer 18

Find a single vertex and rotate that point around the origin.

If z1 is a root, then the roots (vertices) are given by z1, z1ω, z1ω^2,…,z1ω^(n-1) where ω=e^(2πi/n), where n is the number of roots

Question 19

2.1 Describe the method of differences

Answer 19

When summing from 1->n, take one fraction away from another, then sub in r=1 for the first term, then r=2 for the second, and so on, until finally r=n-1 and r=n. Then spot a pattern and see which terms cancel, then simplify

Question 20

2.3 What is the Maclaurin series expansion of the function f(x)? When is it valid?

Answer 20

f(x) = f(0) + f′(0)x + (f′′(0)x^2)/2! + … + (f^(r)(0)x^r)/r! + …

Only valid when f(0), f′(0), f′′(0), …, f^(r)(0), … all have finite values

Question 21

2.4 Which Maclaurin expansions are in the formulae booklet?

Answer 21

e^x, ln(1+x), sinx, cosx, arctanx

Question 22

3.1 What is an improper integral?

Answer 22

• One or both limits is infinite
• f(x) is undefined at x=a or x=b or another point in the interval [a,b]

Question 23

3.2 How do you work out the mean value of a function f(x) over an interval [a,b]?

Answer 23

Mean value of f(x) = (1/(b-a))∫(a->b)f(x) dx

Question 24

3.2 What happens to the mean value of f(x) when undergoing a transformation like af(x)?

Answer 24

The mean value of af(x) over the interval is a times the mean value of f(x) over the same interval

Question 25

3.3 What is the differentiation of arcsinx?

Answer 25

d/dx(arcsinx) = 1/√(1 - x^2)

Question 26

3.3 What is the differentiation of arccosx?

Answer 26

d/dx(arccosx) = -1/√(1 - x^2)

Question 27

3.3 What is the differentiation of arctanx?

Answer 27

d/dx(arctanx) = 1/(1 + x^2)

Question 28

3.4 What is the integral associated with arctan?

Answer 28

∫1/(a^2 + x^2)dx = (1/a)arctan(x/a) + c

Question 29

3.4 What is the integral associated with arcsin?

Answer 29

∫1/√(a^2 - x^2)dx = arcsin(x/a) + c

Question 30

5.1 What are the key points to remember when converting between polar and cartesian coordinates?

Answer 30

x = rcosθ y = rsinθ r^2 = x^2 + y^2 θ = arctan(y/x)

Question 31

5.1 What is the polar form of coordinates?

Answer 31

(r,θ)

Question 32

5.2 What does the curve r=a look like?

Answer 32

A circle with centre O and radius a

Question 33

5.2 What does the curve θ=α look like?

Answer 33

A half-line through O making an angle of α with the initial line (+ve x-axis)

Question 34

5.2 What does the curve r=aθ look like? Where does it cross the axis?

Answer 34

A spiral starting at O Crosses +ve y-axis at aπ/2, -ve x-axis at aπ, -ve y-axis at 3aπ/2, and +ve x-axis at 2aπ

Question 35

5.2 When sketching polar curves with cos(nθ) or sin(nθ), how should you choose the points to plot?

Answer 35

Draw a table of values to check the value of r at multiples of π/2n until you reach 2π/n, ignore any values of r that are negative

Question 36

5.2 What are the shapes of the graph r=a(p + qcosθ)?

Answer 36

- When p=q, a 'cardioid' shape is made with the 'dimple' going through O - When p ≥ 2q, an egg shape - When q ≤ p < 2q, a 'dimple' shape

Question 37

5.3 What is the formula for the area enclosed by a polar curve?

Answer 37

Area = (1/2)(∫A->B(r^2)dθ

Question 38

5.3 What can sin^2 (θ) and cos^2 (θ) also be written as?

Answer 38

sin^2 (θ) = 1/2 - (1/2)cos2θ cos^2 (θ) = 1/2 + (1/2)cos2θ

Question 39

5.4 How do you find the tangent of a polar curve that is parallel to the initial line?

Answer 39

Set dy/dθ = 0

Question 40

5.4 How do you find the tangent of a polar curve that is perpendicular to the initial line?

Answer 40

Set dx/dθ = 0

Question 41

5.4 Prove that the curve r = (p+qcosθ), where p>q, is convex for p≥2q and has a dimple for p<2q

Answer 41

We will check how many vertical tangents there are to the curve so x=rcosθ=pcosθ + qcos^2 θ dx/dθ = -psinθ - 2qsinθcosθ = 0 psinθ + 2qsinθcosθ = 0 sinθ(p + 2qcosθ) = 0 sinθ = 0 or cosθ = -p/2q For the sinθ case, θ = 0, π (2 tangents so far) We will see if we can generate a third tangent. cosθ = -p/2q For p>2q cosθ < -1 so no additional solutions. If p = 2q, cosθ = -1, θ = π which is a repeat so not an additional solution. If q ≤ p < 2q, another solution will be generated (by considering the cosθ = -p/2q) If two tangents are found then the curve is convex and if three tangents are found the curve is a dimple.

Question 42

6.1 Describe the features of sinh

Answer 42

sinhx = (e^x - e^(-x))/2 Domain: x ∈ ℝ Range: f(x) ∈ ℝ Looks like y=x^3, goes through (0,0) Odd function b/c f(-x) = -f(x)

Question 43

6.1 Describe the features of cosh

Answer 43

coshx = (e^x + e^(-x))/2 Domain: x ∈ ℝ Range: f(x) ≥ 1 Looks like y=x^2 with the turning point at (0,1) Even function b/c f(-x) = f(x)

Question 44

6.1 Describe the features of tanh

Answer 44

tanhx = (e^(2x) - 1)/(e^(2x) + 1) or sinhx/coshx Domain: x ∈ ℝ Range: -1 < f(x) < 1 Looks like an s, with horizontal asymptotes at y=-1 and y=1, goes through (0,0)

Question 45

6.1 Describe the features of sech

Answer 45

sechx = 1/coshx = 2/(e^x + e^(-x)) Domain: x ∈ ℝ Range: 0 < f(x) ≤ 1 Has two asymptotes, one on either side of the y-axis, both on line y=0, has a 'bump' in the middle, goes through (0,1)

Question 46

6.1 Describe the features of cosech

Answer 46

cosechx = 1/sinhx = 2/(e^x - e^(-x)) Domain: x ≠ 0 Range: f(x) ≠ 0 Looks like the graph of y=1/x

Question 47

6.1 Describe the features of coth

Answer 47

cothx = 1/tanhx = (e^(2x) + 1)/(e^(2x) - 1) Domain: x ≠ 0 Range: f(x) < -1, f(x) > 1 Looks like y=1/x but instead of horizontal asymptotes at y=0, they are at y=1 in the 1st quadrant and y=-1 in the 3rd quadrant

Question 48

6.2 Describe the features of arsinh

Answer 48

arsinhx = ln(x + √(x^2 + 1)) Domain: x ∈ ℝ Range: f(x) ∈ ℝ Looks like an s shape, through the origin, like y=-x^3 rotated 90 degrees clockwise

Question 49

6.2 Describe the features of arcosh

Answer 49

arcoshx = ln(x + √(x^2 - 1)) Domain: x ≥ 1 Range: f(x) ≥ 0 (Domain of coshx is restricted to x ≥ 0 as it is many-to-one function) Looks like y = √x but starting from (1,0)

Question 50

6.2 Describe the features of artanh

Answer 50

artanhx = (1/2)ln((1+x)/(1-x)) Domain: -1 < x < 1 Range: f(x) ∈ ℝ Looks like y=x^3 but has vertical asymptotes at y=-1 and y=1

Question 51

6.2 Give the domain and range for arsech and describe the shape of the graph

Answer 51

arsechx Domain: 0 < x ≤ 1 Range: f(x) ≥ 0 All within 1st quadrant, has a vertical asymptote at x=0, goes through (1,0), bends both ways

Question 52

6.2 Give the domain and range for arcosech and describe the shape of the graph

Answer 52

arcosechx Domain: x ≠ 0 Range: f(x) ≠ 0 Looks like y=1/x

Question 53

6.2 Give the domain and range for arcoth and describe the shape of the graph

Answer 53

arcothx Domain: x < -1, x > 1 Range: f(x) ≠ 0 Looks like y=1/x but with vertical asymptotes at x=-1 and x=1, the horizontal asymptotes are the same at y=0

Question 54

6.2 Show that arsinhx = ln(x + √(x^2 + 1))

Answer 54

y = arsinhx x = sinhy x = (e^y - e^(-y))/2 2x = e^y - e^(-y) 2xe^y = e^(2y) - 1 e^(2y) - 2xe^y - 1 = 0 Using quadratic formula: e^y = (2x±√(4x^2 + 4))/2 e^y = (2x±(√4)(√(x^2 + 1)))/2 e^y = x±√(x^2 + 1) e^y = x - √(x^2 + 1) can be ignored since √(x^2 + 1) > x, and would give a negative value of e^y, which is not possible y = ln(x+√(x^2 + 1)) So arsinhx = ln(x+√(x^2 + 1))

Question 55

6.2 Show that arcoshx = ln(x + √(x^2 - 1)), x ≥ 1

Answer 55

y = arcoshx x = coshy x = (e^y + e^(-y))/2 2x = e^y + e^(-y) 2xe^y = e^(2y) + 1 e^(2y) - 2xe^y + 1 = 0 Using quadratic formula: e^y = (2x±√(4x^2 - 4))/2 e^y = (2x±(√4)(√(x^2 - 1)))/2 e^y = x±√(x^2 - 1) arcoshx is always non-negative. For all values of x > 1, x - √(x^2 - 1) < 1 so the value of ln(x - √(x^2 -1)) is negative y = ln(x+√(x^2 - 1)) So arcoshx = ln(x+√(x^2 - 1))

Question 56

6.2 Show that artanhx = (1/2)ln((1+x)/(1-x)), -1 < x < 1

Answer 56

y = artanhx x = tanhy x = (e^(2y) - 1)/(e^(2y) + 1) xe^(2y) + x = e^(2y) - 1 xe^(2y) - e^(2y) = - x - 1 (e^(2y))(x - 1) = - (x + 1) e^(2y) = (-(x+1))/(x-1) e^2y = (1+x)/(1-x) 2y = ln((1+x)/(1-x)) y = (1/2)ln((1+x)/(1-x)) So artanhx = (1/2)ln((1+x)/(1-x)), -1 < x < 1 For the domain: x ≤ -1, x ≥ 1, ln((1+x)/(1-x)) is undefined, since (1+x)/(1-x) ≤ 0

Question 57

6.3 What is the hyperbolic identity linking cosh and sinh?

Answer 57

cosh^2 A - sinh^2 A = 1

Question 58

6.3 What are the hyperbolic double angle formulae?

Answer 58

sinh(A±B) = sinhAcoshB ± coshAsinhB cosh(A±B) = coshAcoshB ± sinhAsinhB

Question 59

6.3 What is Osborn's rule for hyperbolic trig? How does this help to determine hyperbolic trig identities?

Answer 59

- Replace cos by cosh: cosA -> coshA - Replace sin by sinh: sinA -> sinhA However: - replace any product of two sin terms by minus the product of the two sinh terms: sinAsinB -> -sinhAsinhB or tan^2 A -> -tanh^2 A You can apply Osborn's rule to any standard trig identities to find the hyperbolic equivalents

Question 60

6.4 What is d/dx(sinhx)?

Answer 60

d/dx(sinhx) = coshx

Question 61

6.4 What is d/dx(coshx)?

Answer 61

d/dx(coshx) = sinhx

Question 62

6.4 What is d/dx(tanhx)?

Answer 62

d/dx(tanhx) = sech^2 x

Question 63

6.4 What is d/dx(arsinhx)?

Answer 63

d/dx(arsinhx) = 1/√(x^2 + 1)

Question 64

6.4 What is d/dx(arcoshx)?

Answer 64

d/dx(arcoshx) = 1/√(x^2 - 1), x > 1

Question 65

6.4 What is d/dx(artanhx)?

Answer 65

d/dx(artanhx) = 1/(1 - x^2), |x|< 1

Question 66

6.5 What is the integral of sinhx?

Answer 66

∫sinhx dx = coshx + c

Question 67

6.5 What is the integral of coshx?

Answer 67

∫coshx dx = sinhx + c

Question 68

6.5 What is the integral of 1/sqrt(x^2 + 1)?

Answer 68

∫1/sqrt(x^2 + 1) dx = arsinhx + c

Question 69

6.5 What is the integral of 1/sqrt(x^2 - 1)?

Answer 69

∫1/sqrt(x^2 - 1) dx = arcoshx + c

Question 70

6.5 What is the integral of sechxtanhx?

Answer 70

∫sechxtanhx dx = -sechx + c

Question 71

6.5 What is the integral of tanhx?

Answer 71

∫tanhx dx = ln(coshx) + c

Question 72

6.5 What is the integral of 1/sqrt(x^2 + a^2)?

Answer 72

∫1/sqrt(x^2 + a^2) dx = arsinh(x/a) + c

Question 73

6.5 What is the integral of 1/sqrt(x^2 - a^2)?

Answer 73

∫1/sqrt(x^2 - a^2) dx = arcosh(x/a) + c

Question 74

6.5 What is the integral of sech^2 x?

Answer 74

∫sech^2 x dx = tanhx + c

Question 75

6.5 What is the integral of sech^2 xtanhx?

Answer 75

∫sech^2 xtanhx dx = (1/2)tanh^2x + c = (-1/2)sech^2 x + c

Question 76

6.5 When integrating 1/sqrt(x^2 - a^2) using substitution, what substitution should you use?

Answer 76

∫1/sqrt(x^2 - a^2) dx Use x = acoshu

Question 77

6.5 When integrating 1/sqrt(x^2 + a^2) using substitution, what substitution should you use?

Answer 77

∫1/sqrt(x^2 + a^2) dx Use x = asinhu

Question 78

6.5 What is the integral of 1/(x^2 + a^2)?

Answer 78

∫1/(x^2 + a^2) dx = (1/a)arctan(x/a) + c

Question 79

6 What is the identity linking tanhx and sechx?

Answer 79

sech^2 x = 1 - tanh^2 x

Question 80

6 How can you write sinh^2 x in terms of cosh?

Answer 80

sinh^2 x = (1/2)cosh2x - 1/2

Question 81

6 How can you write cosh^2 x in terms of cosh?

Answer 81

cosh^2 x = (1/2)cosh2x + 1/2

Question 82

6 What can sinh2x also be written as?

Answer 82

sinh2x = 2sinhxcoshx

Question 83

6 What can cosh2x also be written as?

Answer 83

cosh2x = cosh^2 x + sinh^2 x

Question 84

6 What can tanh2x also be written as?

Answer 84

tanh2x = 2tanhx/(1 + tanh^2 x)

Question 85

7.1 How do you separate the variables of a first-order differential equation of the form dy/dx = f(x)g(y)?

Answer 85

If dy/dx=f(x)g(y), then solve it by ∫(1/g(y))dy = ∫f(x)dx Integrate all the y's on one side and then x's on the other

Question 86

7.1 How do you sketch a family of curves?

Answer 86

If general curve is f(y)=Af(x) then sketch curves with different values of A

Question 87

7.1 How do you use reverse product rule to solve some first-order differential equations?

Answer 87

Look to see if the form f(x)g'(y) + g(y)f'(x) exists somewhere in the equation Then write as d/dx(f(x)g(y)) = RHS Solve by integrating both sides, don't forget to add c

Question 88

7.1 How do you solve first-order differential equations when you can't use the reverse product rule?

Answer 88

- Divide by the coefficient of dy/dx - Integrating factor is e^(∫(coefficient of y)) - Multiply by integrating factor - Solve like a regular first-order differential equation

Question 89

7.2 How do you solve second-order homogenous differential equations?

Answer 89

a(d^2y/dx^2) + b(dy/dx) + cy = 0 Can be written as the auxiliary equation am^2 + bm + c = 0 Auxiliary equation: b^2 - 4ac > 0, 2 distinct roots, general solution is y = Ae^(αx) + Be^(βx) b^2 - 4ac = 0, 1 repeated root, general solution is y = (A + Bx)e^(αx) b^2 - 4ac < 0, no real roots, general solution is y = (e^(px))(Acosqx + Bsinqx) for roots p ± iq

Question 90

7.3 How do you solve second-order non-homogenous differential equations?

Answer 90

a(d^2y/dx^2) + b(dy/dx) + cy = f(x) Find the auxiliary equation as if it was homogenous Solve this equation to find the complimentary function (CF) Choose an appropriate form for the particular integral (PI) and substitute into the original equations and compare coefficients The general solution is y = CF + PI

Question 91

7.3 What form of PI should you use if f(x) is of the form f(x) = k?

Answer 91

f(x) = k PI is f(x) = λ

Question 92

7.3 What form of PI should you use if f(x) is of the form f(x) = ax + b?

Answer 92

f(x) = ax + b PI is f(x) = λx + μ

Question 93

7.3 What form of PI should you use if f(x) is of the form f(x) = ax^2 + bx + c?

Answer 93

f(x) = ax^2 + bx + c PI is f(x) = λx^2 + μx + v

Question 94

7.3 What form of PI should you use if f(x) is of the form f(x) = ke^(px)?

Answer 94

f(x) = ke^(px) PI is f(x) = λe^(px)

Question 95

7.3 What form of PI should you use if f(x) is of the form f(x) = mcos(ax), msin(ax), or mcos(ax) + nsin(ax)?

Answer 95

f(x) = mcos(ax), msin(ax), or mcos(ax) + nsin(ax) PI is f(x) = λcos(ax) + μsin(ax)

Question 96

7.3 When will you need to modify the particular integral?

Answer 96

If term is same as one of the terms in the CF, usually given in the exam question though

Question 97

8.1 How do you solve first-order differential equation modelling problems involving tanks of chemicals?

Answer 97

- Sketch the tank w/ volume of fluid in it, & the volumes coming out & going in - Total volume is V = starting vol. + rate out - rate in - Proportion of desired chemical is x/V - Proportion of desired chemical leaking is the leaking rate multiplied by x/V - Thus the rate of change of amount of chemical is dx/dt = rate in - leaking rate multiplied by x/V

Question 98

8.2 What is simple harmonic motion?

Answer 98

SHM is motion in which acceleration is always towards a fixed point O on line of motion P. Acceleration is proportional to the displacement of P from O ẍ = (-ω^2)x

Question 99

8.2 What is the SHM relationship between acceleration and velocity?

Answer 99

ẍ = v(dv/dt)

Question 100

8.2 What is the time period of the general equation x = Acoskt + Bsinkt?

Answer 100

2π/k

Question 101

8.3 What is damped harmonic motion?

Answer 101

An additional force added to SHM which is proportional to the velocity of the particle. It acts to slow down the particle. ẍ + kẋ + (ω^2)x = 0

Question 102

8.3 What is the first case corresponding to the damped harmonic motion auxiliary equation?

Answer 102

Discriminant > 0: - Two distinct real roots - Heavy damping - No oscillations as resistive force is large compared to restoring force

Question 103

8.3 What is the second case corresponding to the damped harmonic motion auxiliary equation?

Answer 103

Discriminant = 0: - Equal roots - Critical damping - No oscillations

Question 104

8.3 What is the third case corresponding to the damped harmonic motion auxiliary equation?

Answer 104

Discriminant < 0: - Complex roots - Light damping - Oscillations decrease exponentially over time (period of observed oscillations can be calculated)

Question 105

8.3 What is forced harmonic motion?

Answer 105

When a particle is subject to motion like DHM but is also forced to oscillate with frequency other than its natural one ẍ + kẋ + (ω^2)x = f(t) To solve this find the particular integral like in chapter 7